# Strictly increasing bounded function of class $C^1$

Let $$f:\mathbb{R} \to \mathbb{R}$$ be a strictly increasing bounded function of class $$C^1$$. Prove that there exists a sequence $$\{x_n\}_n$$ of real numbers such that $$x_n\to\infty$$ and $$\lim_{n \to \infty} f'(x_n)=0$$.

Then construct a strictly increasing bounded function $$f:\mathbb{R} \to \mathbb{R}$$ of class $$C^1$$ such that the $$\lim_{x \to \infty} f' (x)$$ does not exist.

I know that if we assume $$a_n$$ increases ($$a_{n+1}\ge a_n$$), then either it is bounded, or not. If yes, then it converges to the $$\sup a_n$$, else it goes to $$+\infty$$. Since it is bounded here it goes to a limit and so eventually it approaches the limit and derivative approaches 0. But I dont really know how to prove it rigorously and construct such a function such that limit DNE.

By the Mean Value Theorem, for any $$n\in\mathbb{N}$$ there is $$x_n\in (n,n+1)$$ such that $$f'(x_n)=f(n+1)-f(n).$$ The sequence $$(x_n)_n$$ is strictly increasing and goes to $$+\infty$$. Then $$\lim_{n\to \infty }f'(x_n)=\lim_{n\to \infty }(f(n+1)-f(n))=\lim_{n\to \infty}f(n+1)-\lim_{n\to \infty }f(n)=M-M=0$$ where $$M=\sup\{f(x):x \in \Bbb{R}\}$$.
Hint for the second part. Consider the continuous piecewise function $$f$$ defined in $$\mathbb{R}$$ which is $$e^{x}$$ for $$x\in (-\infty,0]$$, it is linear with slope $$1$$ in each interval $$[n,n+\frac{1}{2^n}]$$ and it is linear with slope $$\frac{1}{2^n}$$ in each interval $$[n+\frac{1}{2^n},n+1]$$ for any $$n\in\mathbb{N}$$. The function $$f$$ is strictly increasing and bounded (why?). Moreover, $$f'(x)$$ attains the value $$1$$ and values $$<1/2$$ infinite times as $$x$$ goes to infinity, and therefore $$\lim_{x \to \infty} f' (x)$$ does not exist. Note that $$f$$ is not $$C^1$$, but it can be made smooth by changing it suitably around the joint points.
• @jamesblack It is $e^x$ just for $x\leq 0$. For $x\geq 0$ on we start from $e^0=1$ and we draw a segments with slope $1$ over $[0,3/2]$, slope $1/2$ over $[3/2,2]$, again slope $1$ over $[2,9/4]$, slope $1/4$ over $[9/4,3]$... Oct 29, 2019 at 17:44
• @jamesblack It is not diffrentiable at the joint point like $3/2,2,9/4$... but with a little effort we can modify a bit the definition of $f$ and have the $C^1$ property. Oct 29, 2019 at 17:47
An increasing and bounded from above function $$f:\Bbb{R} \to \Bbb{R}$$ has always a limit as $$x \to +\infty$$ which is $$L=\sup\{f(x):x \in \Bbb{R}\}$$